Whether the C H bond is stable or the C C bond is stable in the chemical bond

This ......Simpler, right?

1. H-C is a polar bond and C-C is a non-polar bond, which one do you think is stable?

2. H-C is a combination of H nucleus and C-ion (in fact, this is a bit exaggerated, but it is definitely meaningful, because when H nucleus and other atoms share electron pairs, they are basically bare cores, and electron pairs are biased towards other anion atoms, so other atoms look like anions), C-C is two carbon atoms sharing electron pairs, do you think that is stable?

3. Also, methane, diamond and graphite are all in the form of shared electron pairs, or charcoal, is it also c, right? These are all carbon elements, but the ignition point of CH4 (that is, the natural gas used at home) in the air is much lower, which also shows that the H-C bond is relatively weak.

4. There is another class of substances - ethyl chloride, in the substitution reaction of ethane and cl2, the cl atom always replaces the h atom, why? Because the C-C bond in ethane is more stable than the C-H bond, when it hits Cl, it replaces the H atom instead of breaking the C-C bond to form two chloromethanes.

There are other reasons, but it is estimated that no accurate data is available, but these are enough for the landlord to choose to use ......

In terms of their relative relative stability, the C-C bond is more stable than the C-H bond. In general chemical reactions, it is easy to break carbon-hydrogen bonds, but it is difficult to break carbon-carbon bonds. Of course, under some special conditions, the carbon-carbon bond will break without the carbon-hydrogen bond change.

The C-C bond is more stable than the C-H bond, and H is electropositive, which is easy to induce.

b c h8n-nh8 n-h is a polar bond and n-n is a non-polar bond.

Covalent bonds, which are formed by different kinds of atoms, are polar covalent bonds.

The C-H bond is a polar bond.

Theoretically, the electron-absorbing capacity of atoms of different elements is different, so c-h is definitely a polar bond from an electronegativity perspective: x(c)=>x(h)=

That is, carbon's ability to attract electrons is stronger than that of hydrogen.

Hence c-h is a polar bond.

Clause.

First, the shorter the chemical bond, the more stable it is, similar to the ball and stick model, the longer the stick, the more likely it is to have problems.

Clause. Second, the double bond and the triple bond are easily oxidized, and the stability is worse than that of the carbon-carbon single bond, which can be known by looking at the conditions of the reaction.

To put it simply, the stronger the electropositivity of the carbon atom, the easier it is to break the C-H bond, and at the same time, if the generated carbon anion can participate in the conjugation and disperse the positive charge, then the C-H bond of the molecule will be more active.

For example, methane.

The carbon-hydrogen bonds in are very stable. Acetaldehyde.

The hydrogen is easily removed, phenol.

The ortho-para hydrogen of the hydroxyl group is also very active.

The reactivity of the O-H bond is similar to that of the C-H bond. When carbon attached to oxygen has a strong electron-withdrawing effect or a conjugation effect, it loses the positive charge of hydrogen ions, and the hydrogen is easily ionized.

For example, carboxylic acids are electron-withdrawing and conjugated, so they are very acidic. Phenol can be conjugated, so there is a certain acidity.

Well, the source of which search first said so much

If there is still a problem, try to add it is to slow the grip

Electropositivity refers to the ability to lose electrons, which is affected by the covalent bonds between carbon atoms and other functional groups.

The more carbon-carbon bonds, the more active h is, and the stronger the electropositivity of c. The more functional groups, the more affected and the more positively charged the C attached to the functional groups.

Electropositivity of carbon: hexane "benzene" phenol spring "acetic acid" benzoic acid "formic acid.

Hexane has the least functional group, and the special bonds formed between the various carbons in benzene make them active (relative to ethane) and stable;

The rest are more acidic: formic acid is the strongest, phenol is the weakest. Benzoic acid and acetic acid are more active than carboxyl and benzene ring phases, so the acidity is enhanced.

They're usually found in organic matter, so they're not active, and they're generally uncharged.

This is a good question, some secondary school textbooks like to say that isotopes have the same chemical properties, but they are not, especially for H D. The difference in the vibrational energy of the chemical bond is the largest of all stable isotope systems, because the relative mass difference between h and d is doubled. For the example of c-h and c-d, the difference between their zero points can be used, and if the difference of h d in the explicit lease state is not taken into account, the kh kd calculated by using this number is.

This is the upper limit of the Hd rate difference for this kinetic isotopic effect**. In addition to the kinetic isotope effect caused by the zero-point energy, there is also a very important one called quantum mechanical tunneling isotopic effect. This phenomenon refers to the fact that under quantum mechanics, particles are uncertain and exist as a wave function.

This means that a portion of the wave function may exist directly on the other side of the barrier. Therefore, some particles do not need to climb over the energy wall to stimulate the stupid barrier as in classical mechanics, but directly react through this "tunneling effect". This effect is currently only observed on H d, and the tunneling effect of protons is much stronger than that of deuterium.

Therefore, when the reaction is tunneled, the effect on H d is great. <>

Both C H and C D have NMR. The nuclear spin quantum number of the h atom is 1 2, but the nuclear spin quantum number of the d atom is 1. The natural one we usually use is H NMR.

However, if H resonates at 400MHz and D can only resonate at more than 60MHz under the same magnetic field strength, the specific value is not clear. Current NMR technology uses pulse trains, and the frequency of resonance is within a very narrow range. For example, reading the H spectrum is only around 400MHz.

This is because the chemical shift is in the ppm order relative to the resonance frequency. So of course you can't read the signal for d on the h spectrum. And not because d can't have an MRI.

If you want to see D NMR, there is a special D spectrum, which is the same as the F spectrum, C spectrum, B spectrum, P spectrum, etc. I guess maybe the questioner wants to ask about kinetic isotope effects. In the simplest terms, because the masses of h and d are different, the difference is doubled, which leads to the different vibrational energy levels of c h and c d.

C h is higher than C d. In the reaction coordinates of a reaction, such as the reaction of ch bond breakage, the energy level of the process from the substrate to the transition state determines the activation energy of the reaction. Although this reaction is a process of translational breakage of C H bonds, there are still many vibrational energy levels in the substrate ground state and transition state, and the vibrational energy levels are related to the approximate mass, and the difference between the masses of h and d leads to the difference in the vibrational energy levels of the substrate.

If the difference between the vibrational energy levels of the C h and C d bonds in the substrate is the same as the difference between the vibrational energy levels of the C h and C d bonds in the transition state, the kinetic isotope effect is not observed, because it happens to be canceled. However, due to the weakening of the Ch and D bonds in the transition state, the vibrational energy level difference between the C h and C d bonds in the transition state is small, so there is a kinetic isotope effect, and the activation energy of the C H bond will be smaller than that of the C D bond. <>

In the past, I often used isotope tracers to trace and measure kinetic isotopic effects when verifying the reaction mechanism, the reason why isotopes can be used to study the mechanism is inseparable from nuclear magnetic resonance technology, the mass and atomic number of h nuclei are odd, so its self-selected quantum number is 1 2, and the mass of d nuclei is even, the atomic number is odd, and its spin fiber and quantum number is integer 1, which has a non-spherical charge distribution, so it has an electric quadrupole moment, which leads to the widening of the NMR spectral line, which is not conducive to detection. In addition, under the specific magnetic field and debris trace radio frequency conditions, only the resonance signal of one nucleus can be observed, and there is no mixing of different nuclear signals, and the resonance frequency of h is far from that of d, so when the signal of the compound is detected by the h spectrum, the d nucleus cannot find a matching signal on the spectrum. If we replace substrate 1 with the pentadeuterium on the benzene ring, after the beam is cyclized with the alkyne, we will find that the deuterium at the two sites marked in red circle is replaced by 59% h and 77% h.

The reason why isotopes can be used to study the mechanism is inseparable from nuclear magnetic resonance technology, the mass and atomic number of h nuclei are odd, so their self-selected quantum number is 1 2, while the mass of d nuclei is even, the atomic number is odd, and the spin quantum number is integer 1, which has a non-spherical charge distribution, so it has an electric quadrupole moment, which leads to the widening of the NMR spectral line, which is not conducive to detection. In addition, under specific magnetic field and radio frequency conditions, the resonance signal of only one nucleus can be observed, and there is no mixing of different nuclear signals, and the resonance frequency of h is far from that of d, so when the signal of a compound is detected by h spectrum, the nucleus of d d cannot find a matching signal on the spectrum. This is background knowledge.

There are two main applications of isotopes in the study of organic chemistry mechanisms.

1.As a tracer atom, if we change substrate 1 to the pentdeuterium generation on the benzene ring, after cyclization with alkynes, we will find that the deuterium at the two sites marked in the red circle is replaced by 59% H and 77% H. From this we can deduce two conclusions, first:

These two sites are involved in the reaction, and second: the reaction in which these two sites are involved is reversible. From this, we can deduce the following possible reaction mechanism, note that the red and rotten town color circles in the mechanism diagram are the inferences just obtained through isotope tracing.

2.Measuring the change in reaction rate caused by isotope substitutionIn chemical reactions, the difference in the position of the isotope substitution relative to the position of the center of reactivity also causes a difference in the rate of change in the reaction rate, which can be divided into three cases:

1) Primary isotope effects.

2) Secondary isotopic effects.

3) Solvent isotope effect <>

Since the mass of the d atom is twice the mass of the H atom, the vibrational zero energy of the C-D bond is a double of the vibrational zero energy of the C-H bond, which means that in the same potential well, the D atom is in a deeper position; At low temperatures, this affects the rate of reaction of the bikinamine chemistry involved in this bond.

For example, once the h of c-h is replaced by d, it is relatively difficult to replace d back with h, which will lead to the enrichment of c-d relative to c-h at low temperatures. In some astronomical environments, such as inside a cold dark molecular cloud, the temperature is only absolute.

10k), which is an important effect and has been confirmed by many observations. For example, although the abundance of d relative to h in the universe.

Yes, but in some cold dark molecular clouds, the abundance of CH2DOH relative to CH3OH has been observed to reach tens of percent. <>